Homework Help: Basic Atomic Question

1. Apr 19, 2012

ZedCar

1. The problem statement, all variables and given/known data
In question 50 in this link:

http://www.tarleton.edu/Faculty/alow/prob21.htm

I was wondering how the numbers 4.00260 amu + 1.00867 amu – 2.01410 amu – 3.01605 amu been obtained?

Thank you.

2. Relevant equations

3. The attempt at a solution

2. Apr 19, 2012

tiny-tim

3. Apr 20, 2012

ZedCar

Thank you.

The first part of question 50 states 2H + 3H → 4He + 1n

So are they multiplying these numbers by a certain constant to arrive at the numbers of 4.00260 amu + 1.00867 amu – 2.01410 amu – 3.01605 amu

If so, what is this constant?

As the numbers do look just slightly more than the numbers which I have emboldened.

4. Apr 20, 2012

tiny-tim

they're not a constant times anything, atomic weight is not a linear function of the number of nucleons

fuse two protons (1H) together, that's two lots of 1.00867 = 2.01734, you get a 2H (= 2.01410) and some energy left over, which is the energy of fusion that they're trying to make use of in fusion reactors

5. Apr 20, 2012

Staff: Mentor

You may want to change it to something closer to reality

6. Apr 20, 2012

oops!